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Re: GMAT Diagnostic Test Question 40 [#permalink]
13 Jun 2009, 20:54

1

This post was BOOKMARKED

Explanation Official Answer: E

Number of triangles can be formed from a polygon with 7 sides \(= C_7^3 = 35\). Number of quadrilaterals can be formed from a polygon with 7 sides \(= C_7^4 = 35\). Total Number of ways that triangles and rectangles can be formed from a polygon with 7 sides \(= C_7^3 + C_7^4 = 35+35 = 70\). _________________

How do you know that the number of triangles you can make from a 7 sided polygon is 7c3? I guess I'm a bit confused on how these "triangles" are formed.

"Making" triangles boils down to picking 3 different points out of 7 points that the polygon has. This is why it equals \(C_7^3\).

matt0586 wrote:

How do you know that the number of triangles you can make from a 7 sided polygon is 7c3? I guess I'm a bit confused on how these "triangles" are formed.

"Making" triangles boils down to picking 3 different points out of 7 points that the polygon has. This is why it equals \(C_7^3\).

matt0586 wrote:

How do you know that the number of triangles you can make from a 7 sided polygon is 7c3? I guess I'm a bit confused on how these "triangles" are formed.

Having just read the solution the question makes sense. But when I just did the test, the actual question didn't make sense to me. I didn't understand how you could form a triangle from another polygon. I think the question should be worded such that it implies that the vertices of the polygon are the vertices of the triangle/quadrilateral. Just my 2c... if everyone else understands it then

Re: GMAT Diagnostic Test Question 41 [#permalink]
06 Oct 2009, 02:03

I guess so...

'cos, with the question in current wordings, I see the answer to be 13.. but again.. with an assumption that we can draw one line to join three any three points in the given polygon..

dzyubam wrote:

I see your point. Would this rewording be OK:

How many triangles and quadrilaterals altogether can be formed from the vertices of a 7-sided polygon?

Re: GMAT Diagnostic Test Question 41 [#permalink]
23 Dec 2009, 04:34

You'd be right if the question was a probability question. Here you just have to find the number of triangles that can be drawn under these circumstances, the number of quadrilaterals, and finally add up the two numbers. The question doesn't ask neither of number of ways these figures can be organized in pairs nor does it ask of a any probability value.

I hope it helped.

paulnihar5 wrote:

I have a doubt. Isn't "AND" means multiplying and "OR" means adding up?

I ended doing the same calculations except multiplying them instead of adding and never got the correct answer.

I got a question, when i solved the problem, i literary drew 7 sided polygon, then started drawing triangles, now from first point it forms 5 triangles, so MULTIPLYING 5*7 GIVES ANSWER 35. but aren't we suppose to omit repeating results? because I interpret the question as if i am suppose to find number of UNIQUE triangles that can be drawn and it comes out to be 20 instead, and 13 UNIQUE quadrilaterals. if that wasn't meant here, i still think asking to find NO. OF UNIQUE triangles and quadrilaterals formed will help complicate problem even more.

I got a question, when i solved the problem, i literary drew 7 sided polygon, then started drawing triangles, now from first point it forms 5 triangles, so MULTIPLYING 5*7 GIVES ANSWER 35. but aren't we suppose to omit repeating results? because I interpret the question as if i am suppose to find number of UNIQUE triangles that can be drawn and it comes out to be 20 instead, and 13 UNIQUE quadrilaterals. if that wasn't meant here, i still think asking to find NO. OF UNIQUE triangles and quadrilaterals formed will help complicate problem even more.

I don't know how you counted 20 and 13, but \(C^3_7=35\) gives 3 unique points and \(C^4_7=35\) gives 4 unique points, which means that 35 uniques triangles and 35 uniques quadrilaterals can be formed --> 35+35=70. But the answer E (70) to be 100% correct, one thing should be changed in the stem:

Generally in a plane if there are \(n\) points of which no three are collinear, then: 1. The number of triangles that can be formed by joining them is \(C^3_n\).

2. The number of quadrilaterals that can be formed by joining them is \(C^4_n\).

3. The number of polygons with \(k\) sides that can be formed by joining them is \(C^k_n\).

We see that the above formulas are correct if no 3 points are collinear, so we should mention that our 7-sided polygon is not concave. So I'd suggest to change the stem saying: "... using the vertices of a 7-sided regular polygon".

Re: GMAT Diagnostic Test Question 41 [#permalink]
09 Oct 2010, 08:05

basically : 3C7 is the same as 4C7 so the result is quicker calculated by: 2 * 3C7 - it saves few second as there is then only one factorization _________________

Guys, I have a question about the nomenclature used for combinations. In the above quoted text, and in the OA, it says the combination to use is \(C^3_7=35\) . . . . but wouldn't this give us a fraction and lead us to the wrong answer?

I believe it should be \(C^7_3=35\) since

\(C^n_r=\) n!/r!(n-r)!

This would then actually equal 35.

In the OA, the formula would read:

\(C^3_7=35\) which is 3!/7!(-4)! which then equals a fraction and the wrong answer. . .

Where is my misunderstanding? Any help would be greatly appreciated. _________________

Re: GMAT Diagnostic Test Question 41 [#permalink]
01 Oct 2012, 07:18

1

This post received KUDOS

Expert's post

Krest19 wrote:

Bunuel wrote:

NAKAMONIEL wrote:

\(C^3_7=35\)

Guys, I have a question about the nomenclature used for combinations. In the above quoted text, and in the OA, it says the combination to use is \(C^3_7=35\) . . . . but wouldn't this give us a fraction and lead us to the wrong answer?

I believe it should be \(C^7_3=35\) since

\(C^n_r=\) n!/r!(n-r)!

This would then actually equal 35.

In the OA, the formula would read:

\(C^3_7=35\) which is 3!/7!(-4)! which then equals a fraction and the wrong answer. . .

Where is my misunderstanding? Any help would be greatly appreciated.

\(C^3_7\) and \(C^7_3\) are just different ways of writing the same: choosing 3 out of 7.

Re: GMAT Diagnostic Test Question 40 [#permalink]
08 Jan 2014, 15:23

GMAT TIGER wrote:

Explanation Official Answer: E

Number of triangles can be formed from a polygon with 7 sides \(= C_7^3 = 35\). Number of quadrilaterals can be formed from a polygon with 7 sides \(= C_7^4 = 35\). Total Number of ways that triangles and rectangles can be formed from a polygon with 7 sides \(= C_7^3 + C_7^4 = 35+35 = 70\).

how do you know we have to use combination here?? is there any other methods to sole this problem? thanks in advance..

Re: GMAT Diagnostic Test Question 40 [#permalink]
09 Feb 2014, 15:09

C^4_7=35 makes sense to me in theory, but when I count the number quadrilaterals formed at each vertical, I only find four. Can someone please help me picture the math visually?

Re: GMAT Diagnostic Test Question 40 [#permalink]
09 Feb 2014, 23:56

Expert's post

Catcat wrote:

C^4_7=35 makes sense to me in theory, but when I count the number quadrilaterals formed at each vertical, I only find four. Can someone please help me picture the math visually?

How many triangles and quadrilaterals altogether can be formed using the vertices of a 7-sided regular polygon?

A. 35 B. 40 C. 50 D. 65 E. 70

Any 3 vertices from 7 can form a triangle and any 4 vertices from 7 can form a quadrilateral, so total of \(C^3_7+C^4_7=35+35=70\) different triangles and quadrilaterals can be formed.